Antithesis: There are integersnnn such that an≠bnsubscriptansubscriptbna_{n}\neq b_{n}; let νν\nu (≧0absent0\geqq 0) be least of them.

We can choose from the point set an infinitesequencez1,z2,z3,…subscriptz1subscriptz2subscriptz3normal-…z_{1},\,z_{2},\,z_{3},\,\ldots which converges to z0subscriptz0z_{0} with zn≠z0subscriptznsubscriptz0z_{n}\neq z_{0} for every nnn. Let zzz in the equation (1) belong to
{z1,z2,z3,…}subscriptz1subscriptz2subscriptz3normal-…\{z_{1},\,z_{2},\,z_{3},\,\ldots\} and let’s divide both sides of (1) by (z-z0)νsuperscriptzsubscriptz0ν(z-z_{0})^{{\nu}} which is distinct from zero; we then have

Let here zzz to tend z0subscriptz0z_{0} along the points z1,z2,z3,…subscriptz1subscriptz2subscriptz3normal-…z_{1},\,z_{2},\,z_{3},\,\ldots, i.e. we take the limitslimn→∞⁡f⁢(zn)subscriptnormal-→nfsubscriptzn\lim_{{n\to\infty}}f(z_{n}) and limn→∞⁡g⁢(zn)subscriptnormal-→ngsubscriptzn\lim_{{n\to\infty}}g(z_{n}). Because the sum of powerseries is always a continuous function, we see that in (2),

But all the time, the left and right side of (2) are equal, and thus also the limits. So we must have aν=bνsubscriptaνsubscriptbνa_{{\nu}}=b_{{\nu}}, contrary to the antithesis. We conclude that the antithesis is wrong. This settles the proof.